Lecture 1: stack containing min function

subject ：AcWing 41. contain min Function of the stack
Design a support push,pop,top Wait for operation and can be in O(1) Retrieve the stack of the smallest elements in time .

push(x)– Put the element x Insert into the stack
pop()– Remove the top element
top()– Get the stack top element
getMin()– Get the smallest element in the stack
Data range
Total number of operation commands [0,100].

Examples

MinStack minStack = new MinStack();
minStack.push(-1);
minStack.push(3);
minStack.push(-4);
minStack.getMin(); --> Returns -4.
minStack.pop();
minStack.top(); --> Returns 3.
minStack.getMin(); --> Returns -1.

Topic analysis ：
If this problem does not return the minimum value, then we can directly use the stack , What if you need to return the minimum value ？
We can use auxiliary stack
The sub group stack only stores elements smaller than the top element , Then the top of the auxiliary stack is the minimum value of the current stack .
According to the following figure, we can intuitively feel the relationship between stack and auxiliary stack .
Even if 3 Pop up stack , that 41 Will become the smallest value of the current stack .

class MinStack {
    
public:
    /** initialize your data structure here. */
    MinStack() {
    
        
    }
    
    void push(int x) {
    
        stk.push(x);
        if(min_stk.empty()||min_stk.top()>=x)min_stk.push(x);
    }
    
    void pop() {
    
        if(min_stk.top()==stk.top())min_stk.pop();
        stk.pop();
    }
    
    int top() {
    
        return stk.top();
    }
    
    int getMin() {
    
        return min_stk.top();
    }
private:
    stack<int> stk,min_stk;
};

/** * Your MinStack object will be instantiated and called as such: * MinStack obj = new MinStack(); * obj.push(x); * obj.pop(); * int param_3 = obj.top(); * int param_4 = obj.getMin(); */